定义： a,b,c是整数，且ab!=0，那么形如ax+by=c的方程称为二元一次不定方程
定理1： 设a,b是整数且 d = gcd(a,b) 如果d|c，那么返程存在无穷多个整数解，否则不存在整数解。
这里大概可以发现二元一次不定方程和同余方程可以相互转化，如在a>0且b>0的条件下，求二元一次方程ax+by=c整数解等价于求一元线性同余方程ax≡c(mod b）的整数解。
给出poj 2142这个题练练手
The Balance
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 5225 | Accepted: 2297 |
Description
Ms. Iyo Kiffa-Australis has a balance and only two kinds of weights to measure a dose of medicine. For example, to measure 200mg of aspirin using 300mg weights and 700mg weights, she can put one 700mg weight on the side of the medicine and three 300mg weights on the opposite side (Figure 1). Although she could put four 300mg weights on the medicine side and two 700mg weights on the other (Figure 2), she would not choose this solution because it is less convenient to use more weights.
You are asked to help her by calculating how many weights are required. 
Input
The input is a sequence of datasets. A dataset is a line containing three positive integers a, b, and d separated by a space. The following relations hold: a != b, a <= 10000, b <= 10000, and d <= 50000. You may assume that it is possible to measure d mg using a combination of a mg and b mg weights. In other words, you need not consider "no solution" cases.
The end of the input is indicated by a line containing three zeros separated by a space. It is not a dataset.
Output
The output should be composed of lines, each corresponding to an input dataset (a, b, d). An output line should contain two nonnegative integers x and y separated by a space. They should satisfy the following three conditions.
- You can measure dmg using x many amg weights and y many bmg weights.
- The total number of weights (x + y) is the smallest among those pairs of nonnegative integers satisfying the previous condition.
- The total mass of weights (ax + by) is the smallest among those pairs of nonnegative integers satisfying the previous two conditions.
No extra characters (e.g. extra spaces) should appear in the output.
Sample Input
700 300 200 500 200 300 500 200 500 275 110 330 275 110 385 648 375 4002 3 1 10000 0 0 0
Sample Output
1 3 1 1 1 0 0 3 1 1 49 74 3333 1
Source
用扩展欧几里的解决问题。
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4
5 using namespace std;
6
7 int gcd(int a,int b)
8 {
9 if(b == 0) return a;
10 else return gcd(b,a%b);
11 }
12
13
14 int ex_gcd(int a,int b,int &x,int &y)
15 {
16 if(b == 0){
17 x = 1;
18 y = 0;
19 return a;
20 }
21 else {
22 ex_gcd(b,a%b,x,y);
23 int t = x;
24 x = y;
25 y = t - a/b*y;
26 return t;
27 }
28 }
29
30 int main()
31 {
32 int a,b,c,x,y;
33 while(scanf("%d%d%d",&a,&b,&c)!=EOF){
34 if(a == 0 && b == 0 && c == 0) break;
35 int d = gcd(a,b);
36 a /= d;
37 b /= d;
38 c /= d;
39 ex_gcd(a,b,x,y);
40 int xx = x*c;
41 xx = (xx%b+b)%b;
42 int yy = (c - a*xx)/b;
43 if(yy < 0) yy = -yy;
44 y = y*c;
45 y = (y%a+a)%a;
46 x = (c - b*y)/a;
47 if(x < 0) x = -x;
48 if(x+y > xx+yy){
49 x = xx;
50 y = yy;
51 }
52 printf("%d %d\n",x,y);
53 }
54
55 return 0;